5

This seems extremely trivial but I want to make really sure so I'm posting it. This is a yes or no question.. (Sorry for posting this kind of question but I really wonder if I think wrong)

Here is a Gauss's lemma (Dummit&Foote version)

Let $R$ be a UFD and $F$ be the field of quotients of $R$.

Let $A,B\in F[X]$ be nonconstant polynomials such that $AB\in R[X]$. Then, there are nonzero elements $r,s\in F$ such that $rA, sB$ both lie in $R[X]$ and $AB=(rA)(sB)$.

Is this statement equivalent to the below statement?

Let $f,g\in F[X]$ be nonzero polynomials such that $fg\in R[X]$. Then, there exists $r\in F\setminus\{0\}$ such that $rf$ and $\frac{1}{r} g$ both lie in $R[X]$.

(I'm asking this since I don't get why Dummit&Foote wrote two symbols $r,s$ rather than $r,\frac{1}{r}$)

Community
  • 1
Rubertos
  • 12,491

1 Answers1

2

Yes, this is correct, notice that $$p(x) = A(x)B(x) = a(x)b(x) = \big(rA(x)\big)\big(sB(x)\big) = rsA(x)B(x)$$ and since a UFD ($F[x]$ in this case, which is actually an ED since $F$ is a field) is an integral domain, we can cancel out to get $rs = 1$.

Robert Cardona
  • 9,271